Problem: Simplify and expand the following expression: $ \dfrac{4}{4k + 36}- \dfrac{2}{k + 1}+ \dfrac{1}{k^2 + 10k + 9} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{4}{4k + 36} = \dfrac{4}{4(k + 9)}$ We can factor the quadratic in the third term: $ \dfrac{1}{k^2 + 10k + 9} = \dfrac{1}{(k + 9)(k + 1)}$ Now we have: $ \dfrac{4}{4(k + 9)}- \dfrac{2}{k + 1}+ \dfrac{1}{(k + 9)(k + 1)} $ The least common multiple of the denominators is: $ 4(k + 9)(k + 1)$ In order to get the first term over $4(k + 9)(k + 1)$ , multiply by $\dfrac{k + 1}{k + 1}$ $ \dfrac{4}{4(k + 9)} \times \dfrac{k + 1}{k + 1} = \dfrac{4(k + 1)}{4(k + 9)(k + 1)} $ In order to get the second term over $4(k + 9)(k + 1)$ , multiply by $\dfrac{4(k + 9)}{4(k + 9)}$ $ \dfrac{2}{k + 1} \times \dfrac{4(k + 9)}{4(k + 9)} = \dfrac{8(k + 9)}{4(k + 9)(k + 1)} $ In order to get the third term over $4(k + 9)(k + 1)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{1}{(k + 9)(k + 1)} \times \dfrac{4}{4} = \dfrac{4}{4(k + 9)(k + 1)} $ Now we have: $ \dfrac{4(k + 1)}{4(k + 9)(k + 1)} - \dfrac{8(k + 9)}{4(k + 9)(k + 1)} + \dfrac{4}{4(k + 9)(k + 1)} $ $ = \dfrac{ 4(k + 1) - 8(k + 9) + 4} {4(k + 9)(k + 1)} $ Expand: $ = \dfrac{4k + 4 - 8k - 72 + 4}{4k^2 + 40k + 36} $ $ = \dfrac{-4k - 64}{4k^2 + 40k + 36}$ Simplify: $ = \dfrac{-k - 16}{k^2 + 10k + 9}$